In-the-sequence-of-numbers-1-2-11-22-111-222-the-sum-of-the-digits-in-999th-terms-is-

Question Number 42494 by Tawa1 last updated on 26/Aug/18

In the sequence of numbers 1, 2, 11, 22, 111, 222, \dots the sum of the digits

in 999 th terms is ? ?

Commented by Tinku Tara last updated on 22/Apr/21

1 st term 1

3 rd term 11

5 th term 111

7 th term 1111

9 th term 11111

(2 n + 1) term (n + 1) time 1 .

999 = 2 \times 499 + 1

999 term will have 500 1 s

and sum of digits = 500

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Aug/18

t h e g i v e n s e r i e s i s c o m b i n a t i o n o f t w o s e r i e s

s e r i d s a r e (1, 11, 111, \dots) a n d

(2, 22, 222, \dots .)

f i r s t s e r i e s (1, 11, 111, 1111 \dots)

{(10^{0} + 1), (10^{1} + 1), (10^{2} + 10^{2} + 1), (10^{3} + 10^{2} + 10 + 1) \dots}

999 t h t e r m

10^{999 - 1} + 10^{997} + 10^{996} + \dots + 1 = T_{999}

T_{999} = \frac{10^{998} (1 - \frac{1}{10^{999}})}{1 - \frac{1}{10}} = \frac{10^{999} (1 - \frac{1}{10^{999}})}{9}

T_{999} = \frac{10^{999} - 1}{9} = f (10)

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18

c h e c k t h e a n s w e r

Commented by Tawa1 last updated on 27/Aug/18

Oh, God bless you sir . I {don}^{'} t know the answer . I will just write your

solution sir . is the answer whole number or fraction if you solve

further

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